a) \(\left(x-1\right)\left(2x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\Rightarrow x=1\\2x-4=0\Rightarrow x=2\end{matrix}\right.\)

b) \(\left(x^2+5\right)\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x-5=0\Rightarrow x=5\end{matrix}\right.\)

mà \(x\in Z\Rightarrow x=5\)

c) \(\left(x^2+5\right)\left(x^2-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x^2-2=0\Rightarrow x=\sqrt{2}\end{matrix}\right.\)

mà \(x\in Z\Rightarrow x\in\varnothing\)

\(a.\left(x-4\right)\left(x+7\right)=0\)

\(\Rightarrow\hept{\begin{cases}x-4=0\\x+7=0\end{cases}\Rightarrow\hept{\begin{cases}x=4\\x=-7\end{cases}}}\)

\(b.x\left(x+3\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\x+3=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=-3\end{cases}}}\)

\(c.\left(x-2\right)\left(5-x\right)=0\)

\(\Rightarrow\hept{\begin{cases}x-2=0\\5-x=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\x=5\end{cases}}}\)

\(d.\left(x-1\right)\left(x^2+1\right)=0\)

\(\Rightarrow\hept{\begin{cases}x-1=0\\x^2+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\x^2=-1\end{cases}\Rightarrow}\hept{\begin{cases}x=1\\x=-\left(-1\right)or\left(-1\right)\end{cases}}}\)

a) ( x - 4 ) . ( x + 7 ) = 0

một phép nhân có tích bằng 0 

=> một trong hai thừa số này bằng 0 

+) nếu x - 4 = 0 => x = 0 + 4 = 4

+) nếu x + 7 = 0 => x = 0 - 7 = -7

vậy x = { 4 ; -7 }

b) x . ( x + 3 ) = 0

x + 3 = 0 : x

x + 3 = 0

x = 0 - 3

x = -3

vậy x = -3

c) ( x - 2 ) . ( 5 - x ) = 0

một phép nhân có tích bằng 0 

=> một trong hai thừa số này bằng 0 

+) nếu x - 2 = 0 => x = 0 + 2 = 2

+) nếu 5 - x = 0 => x = 5 - 0 = 5

vậy x = { 2 ; 5 }

d) ( x - 1 ) . ( x² + 1 ) = 0

=> x - 1 = 0 hoặc x² + 1 = 0

+) x - 1 = 0 => x = 0 + 1 = 1

+) x² + 1 = 0 => x² = 0 - 1 = -1 => x = -1

vậy x = { 1 ; -1 }

a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)

Th1 : \(x-\dfrac{1}{2}=0\)

         \(x=0+\dfrac{1}{2}\)

         \(x=\dfrac{1}{2}\)

Th2 : \(-3-\dfrac{x}{2}=0\)

         \(\dfrac{x}{2}=-3\)

         \(x=\left(-3\right)\cdot2\)

         \(x=-6\)

Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)

b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)

    \(x=\dfrac{5}{8}+\dfrac{1}{8}\)

   \(x=\dfrac{3}{4}\)

c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\)

                \(\dfrac{3}{2}+x=-\dfrac{1}{2}-\left(-2\right)\)

                \(\dfrac{3}{2}+x=\dfrac{3}{2}\)

                       \(x=\dfrac{3}{2}-\dfrac{3}{2}\)

                      \(x=0\)

d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)

    \(x+\dfrac{1}{3}=-4\)

    \(x=-4-\dfrac{1}{3}\)

    \(x=-\dfrac{13}{3}\)

a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)

f) \(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)

g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)

i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)

\(E=\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)

\(A=\left\{1;-4\right\}\)

\(B=\left\{2;-1\right\}\)

a) Với mọi x thuộc A đều thuộc E \(\Rightarrow A\subset E\)

Với mọi x thuộc B đều thuộc E \(\Rightarrow B\subset E\)

b) \(A\cap B=\varnothing\)

\(\Rightarrow E\backslash\left(A\cap B\right)=\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)

\(A\cup B=\left\{-4;-1;1;2\right\}\)

\(\Rightarrow E\backslash\left(A\cup B\right)=\left\{-5;-3;-2;0;3;4;5\right\}\)

\(\Rightarrow E\backslash\left(A\cup B\right)\subset E\backslash\left(A\cap B\right)\)

a) \(\left(x-4\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-4=0\\x-7=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=7\end{array}\right.\)

b) \(x\left(x+3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-3\end{array}\right.\)

c) \(\left(x-2\right)\left(5-x\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\5-x=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=5\end{array}\right.\)

d) \(\left(x-1\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x-1=0\) ( Vì \(x^2+1>0\) )

\(\Leftrightarrow x=1\)

a)

\(\left(x-4\right)\left(x-7\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\x=7\end{array}\right.\)

Vậy x = 4 ; x = 7

b)

\(x\left(x+3\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=-3\end{array}\right.\)

Vậy x = 0 ; x = - 3

c)

\(\left(x-2\right)\left(5-x\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=5\end{array}\right.\)

Vậy x = 2 ; x = 5

d)

\(\left(x-1\right)\left(x^2+1\right)=0\)

Mà \(x^2+1\ge1\)

=> x = - 1

Vậy x = - 1

a, \(\left(x-4\right).\left(x-7\right)=0\)

\(\Rightarrow x-4=0\) hoặc \(x-7=0\)

+) \(x-4=0\Rightarrow x=4\)

+) \(x-7=0\Rightarrow x=7\)

Vậy x = 4 hoặc x = 7

b, \(x.\left(x+3\right)=0\)

\(\Rightarrow x=0\) hoặc \(x+3=0\)

+) \(x+3=0\Rightarrow x=-3\)

Vậy x = 0 hoặc x = -3

c, \(\left(x-2\right).\left(5-x\right)=0\)

\(\Rightarrow x-2=0\) hoặc \(5-x=0\)

+) \(x-2=0\Rightarrow x=2\)

+) \(5-x=0\Rightarrow x=5\)

Vậy x = 2 hoặc x = 5

d, \(\left(x-1\right).\left(x^2+1\right)=0\)

\(\Rightarrow x-1=0\) hoặc \(x^2+1=0\)

+) \(x-1=0\Rightarrow x=1\)

+) \(x^2+1=0\Rightarrow x^2=-1\Rightarrow\) không có giá trị x thỏa mãn đề bài.

Vậy x = 1 hoặc không có giá trị x thỏa mãn đề bài

\(8,1-\left(x-6\right)=4\left(2-2x\right)\)

\(\Leftrightarrow1-x+6=8-8x\)

\(\Leftrightarrow-x+8x=8-1-6\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

\(9,\left(3x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)

\(10,\left(x+3\right)\left(x^2+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)

`8)1-(x-5)=4(2-2x)`

`<=>1-x+5=8-6x`

`<=>5x=2<=>x=2/5`

`9)(3x-2)(x+5)=0`

`<=>[(x=2/3),(x=-5):}`

`10)(x+3)(x^2+2)=0`

  Mà `x^2+2 > 0 AA x`

 `=>x+3=0`

`<=>x=-3`

`11)(5x-1)(x^2-9)=0`

`<=>(5x-1)(x-3)(x+3)=0`

`<=>[(x=1/5),(x=3),(x=-3):}`

`12)x(x-3)+3(x-3)=0`

`<=>(x-3)(x+3)=0`

`<=>[(x=3),(x=-3):}`

`13)x(x-5)-4x+20=0`

`<=>x(x-5)-4(x-5)=0`

`<=>(x-5)(x-4)=0`

`<=>[(x=5),(x=4):}`

`14)x^2+4x-5=0`

`<=>x^2+5x-x-5=0`

`<=>(x+5)(x-1)=0`

`<=>[(x=-5),(x=1):}`

\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

Rảnh rỗi thật sự .-.